\documentclass[11pt]{article}
%Gummi|063|=)
\title{\textbf{Assignment 1}}

\date{}
\begin{document}

\maketitle

\setcounter{section}{2}
\section{Give an example which shows that pairwise independence does not imply mutial independence}


\begin{tabular}{|l|l|l|l|}
\hline
	$x_1$ & $x_2$ & $x_3$ & p\\
\hline
	0 & 0 & 0 & 1/4\\
\hline
	0 & 0 & 1 & 0\\
\hline
	0 & 1 & 0 & 0\\
\hline
	0 & 1 & 1 & 1/4\\
\hline
	1 & 0 & 0 & 0\\
\hline
	1 & 0 & 1 & 1/4\\
\hline
	1 & 1 & 0 & 1/4\\
\hline
	1 & 1 & 1 & 0\\
\hline
\end{tabular}


Hence:

\begin{tabular}{|l|l|l|}
\hline
	$x_1$ & $x_2$ & p\\
\hline
	0 & 0 & 1/4\\
\hline
	0 & 1 & 1/4\\
\hline
	1 & 0 & 1/4\\
\hline
	1 & 1 & 1/4\\
\hline
\end{tabular}

and 

\begin{tabular}{|l|l|}
\hline
	$x_1$ & p\\
\hline
	0 & 1/2\\
\hline
	1 & 1/2\\
\hline
\end{tabular}
\\
p($x_1$)=p($x_2$)=p($x_3$)=1/2 \\
p($x_1$,$x_2$)=p($x_2$,$x_3$)=p($x_1$,$x_3$)=1/2=p($x_1$)p($x_2$)=p($x_2$)p($x_3$)=p($x_1$)p($x_3$) \\ 
But joint distribution p($x_1$,$x_2$,$x_3$) $\neq$ p($x_1$)p($x_2$)p($x_3$)=1/8

\end{document}
